An Olympiad geometry question on area lemma. In ABC, E, F, G are points on AB, BC, CA respectively such that $\frac{AE}{EB} = \frac{BF}{FC} = \frac{CG}{GA} = \frac{1}{3}$. K, L, M are the intersection points of the lines AF and CE, BG and AF, CE and BG, respectively. Suppose the area of ABC is 1; find the area of KLM.
I know there is a general formula when dividing in m:n ie. $\frac{(m-n)^2}{m^2+mn+n^2}$
P.S. please no Bary/ Complex
To verify the result, I put $m=n$ and it worked. Also, I tried the area lemma, Ceva's theorem but to no use. I tried this problem for a long time and then realised that Menelaus worked but as this excercise was in the area and ratio section, that felt like 'cheating' as I want to know the author's intended method.
Thanks!
 A: 
Let [.] denote areas. Evaluate the area ratio
$$\frac{[CBM]}{[EBM]}= \frac{[CBG]}{[EBG]}
= \frac{\frac 14 [ABC]}{\frac 34 [ABG]}= \frac{\frac 14 [ABC]}{\frac 34 \frac 34[ABC]}
=\frac 49 $$
which leads to the area $[CBM]$, and likewise $[ACK]$, $[BAL]$
$$[CBM] = \frac{4}{4+9} [CEB] = \frac{4}{13}\frac 34[ABC]
=\frac{3}{13} = [ACK] = [BAL] $$
Then
$$[KLM] = [ABC] -[CBM]- [ACK]-[BAL] =1-3\cdot\frac{3}{13}=\frac 4{13}$$

The derivation applies to general ratio $m:n$ as well
$$\frac{[CBM]}{[EBM]}=\frac{m(m+n)}{n^2}\implies 
\frac{[CBM]}{[CBE]}=\frac{m(m+n)}{m(m+n)+n^2} $$
$$\implies [CBM]= \frac{m(m+n)}{m(m+n)+n^2}\cdot\frac{n}{m+n}
=\frac{m n}{m^2+mn +n^2}$$
$$[KLM] = 1-3[CBM]= \frac{(m-n)^2}{m^2+mn +n^2}
$$
A: The general case of this problem is called Routh's theorem.  It states that if $EB/AE = x$, $FC/BF = y$, $GA/CG = z$, then $$\frac{[KLM]}{[ABC]} = \frac{(xyz-1)^2}{(xy+y+1)(yz+z+1)(zx+x+1)}.$$  Note I have taken the reciprocals so that in your case, $x = y = z = 3$.  The proof of this theorem is given in the Wikipedia link.
